Suppose we have a sample space: S={E_1,E_2,E_3,E_4,E_5,E_6,E_7 }, with probabilities: P(E_1)=0.05, P(E_2)=0.20, P(E_3)=0.20, P(E_4)=0.25, P(E_5)=0.15, P(E_6)=0.10,...

Suppose we have a sample space: S={E_1,E_2,E_3,E_4,E_5,E_6,E_7 }, with probabilities: P(E_1)=0.05, P(E_2)=0.20, P(E_3)=0.20, P(E_4)=0.25, P(E_5)=0.15, P(E_6)=0.10, and P(E_7)=0.05. Consider the following events with their corresponding set of sample points: A={E_1,E_4,E_6 }, B={E_2,E_4,E_7 }, and C={E_2,E_3,E_5,E_7 }. Note: P(A∪B)=P(A)+P(B)-P(A∩B) P(A∪B)=P(A)+P(B) if A and B are mutually exclusive P(A∩B)=0 if A and B are mutually exclusive P(A^C)=1-P(A) P(A∣B)=(P(A∩B))/(P(B)) P(A∣B)=P(A) if A and B are independent Questions: Find the probabilities: P(A), P(B) and P(C). Find the set of sample points of events A∪B (this is a list of sample points {…}) and find P(A∪B) ? Find the set of sample points of A∩B and find P(A∩B). Now use the formula P(A∪B)=P(A)+P(B)-P(A∩B) to confirm your answer from part (b). Are events A and C mutually exclusive? Find B^c and P(B^c) Calculate P(A∣B) and P(B∣A). Are they the same? Are events A and B independent?

Expert Solution

Consider the given problem here “A = {E1, E4, E6}”, => P(A) = P(E1)+P(E4)+P(E6) = 0.05+0.25+0.1 = 0.4, => P(A)=0.4. Similarly, “B = {E2, E4, E7}”, => P(B) = P(E2)+P(E4)+P(E7) = 0.2+0.25+0.05 = 0.5, => P(B)=0.5. Finally, “C = {E2, E3, E5 E7}”, => P(C) = P(E2)+P(E3)+P(E5) +P(E7) = 0.2+0.2+0.15 +0.05 = 0.6, => P(C)=0.6.

Now, “(AUB) = {E1, E4, E6, E2, E7}”, be the sample space. So, “P(AUB)” is given by.

=> P(AUB) = P(E1) + P(E4) + P(E6) + P(E2) + P(E7) = 0.05 + 0.25 + 0.1 + 0.2 + 0.05 = 0.65, => P(AUB) = 0.65”.

Now, “(AUB) = A + B - AB = {E1, E4, E6} + {E2, E4, E7} - {E4} = {E1, E4, E6, E2, E7}. So, “P(AUB)” is given by, where “AB = A intersection B”.

=> P(AUB) = P(A) + P(B) – P(AB) = 0.4 + 0.5 – P(E4) = 0.9 – 0.25 = 0.65, => P(AUB) = 0.65”.

Now, “A = {E1, E4, E6}” and “C = {E2, E3, E5 E7}”, => “AC = {0}”, => “A” and “C” are mutually exclusive events.

Now, “B = {E2, E4, E7}” and the total sample space is given by, “S = {E1, E2, E3, E4, E5, E6, E7}”, => B^c = S-B = {E1, E3, E5, E6}, “B^c = {E1, E3, E5, E6}”. So, “P(B^c) = P(E1) + P(E3) + P(E5) + P(E6) = 0.05 + 0.2 + 0.15 + 0.1 = 0.5, => P(B^c) = 0.5”.

Now, the “P(A|B) = P(AB)/P(B) = 0.25/0.5 = 0.5” and “P(B|A) = P(AB)/P(A) = 0.25/0.4 = 0.625”.

Now, the “P(A)=0.4” and “P(B)=0.5”, => “P(A)*P(B)=0.4*0.5 = 0.2” and “P(AB) = 0.25”. Now, “P(AB)” and “P(A)*P(B)” are not equal, => they are not independent.

Rahul Sunny answered 2 years ago

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